Question

Show that $\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\sin A+\cos A$

Answer 1

Let usfirstfind the value of left hand side (LHS) that is $\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}$ as shown below:

$\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}$

$=\frac{\cos A}{1-\frac{\sin A}{\cos A}}+\frac{\sin A}{1-\frac{\cos A}{\sin A}}(\because \cot x=\frac{\cos x}{\sin x},\tan x=\frac{\sin x}{\cos x})$

$=\frac{\cos A}{\frac{\cos A-\sin A}{\cos A}}+\frac{\sin A}{\frac{\sin A-\cos A}{\sin A}}$
$=\frac{\left(\cos A\right)\left(\cos A\right)}{\cos A-\sin A}+\frac{\left(\sin A)\right(\sin A)}{\sin A-\cos A}$

$=\frac{{\cos }^{2}A}{\cos A-\sin A}-\frac{{\sin }^{2}A}{\cos A-\sin A}$
$=\frac{{\cos }^{2}A-{\sin }^{2}A}{\cos A-\sin A}$
$=\frac{(\cos A-\sin A)(\cos A+\sin A)}{\cos A-\sin A}(\because a^2-b^2=(a+b\left)\right(a-b\left)\right)$
$=\cos A+\sin A=RHS$

Since $LHS=RHS,$

Hence, $\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\cos A+\sin A$.